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# Test: Thermodynamics of Electrochemistry

## 24 Questions MCQ Test Chemistry Class 12 | Test: Thermodynamics of Electrochemistry

Description
This mock test of Test: Thermodynamics of Electrochemistry for JEE helps you for every JEE entrance exam. This contains 24 Multiple Choice Questions for JEE Test: Thermodynamics of Electrochemistry (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Thermodynamics of Electrochemistry quiz give you a good mix of easy questions and tough questions. JEE students definitely take this Test: Thermodynamics of Electrochemistry exercise for a better result in the exam. You can find other Test: Thermodynamics of Electrochemistry extra questions, long questions & short questions for JEE on EduRev as well by searching above.
QUESTION: 1

### Only One Option Correct Type This section contains 15 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE is correct Q. Temperature coefficient of EMF of a cell in terms of entropy change is

Solution:

is called temperature coefficient of EMF of a cell. Thus,
temperature coefficient of

QUESTION: 2

Solution:

QUESTION: 3

### Given ,   Thus ,equilibrium constant for the reaction in terms of log k is 2Fe3+ + 3I-  2Fe2+ + I-3

Solution:

QUESTION: 4

Given ,

Thus ,(log Keq) for the reaction   Cu2+ +In2+  Cu+ + In3+ is

Solution:

QUESTION: 5

An excess of liquid mercury is added to an acidified solution of 1.0 x 10-3 M Fe3+ .Thus  is if 5% of Fe3+ remains at equilibrium at 298 K

Solution:

When equilibrium is set up

QUESTION: 6

EMF of the following cell is 0.2905 V

Zn/Zn2+ (a = 0.1M)|| Fe2+ (a = 0.01M)| Fe

The equilibrium constant for the cell reaction is

[IIT JEE 2004]

Solution:

QUESTION: 7

The half-cell reactions for rusting of iron are

ΔG° (in kJ) for the reaction is

[IIT JEE 2005]

Solution:

In rusting of iron, Fe2+ is formed. Thus

QUESTION: 8

The Gibbs free energy for the decomposition of Al2O3 at 800 K is as follows :

2Al2O→ 4Al + 3O2 , ΔrG = 2898kJ mol-1

The potential difference needed for electrolytic reduction of Al2O3 is at least

Solution:

n= electrons exchanged = 12 in the given reaction

QUESTION: 9

Given ,

The value of standard electrode potential for the half-reaction is

Fe3+(aq) + e- → Fe2+(aq)

Solution:

Since, different number of electrons are involved hence

QUESTION: 10

For the reaction, 2H2(g) + O2(g) → 2H2O(l), E0cell = 1.23 V at 298 K

and  ΔH0(H2O) = - 285.8 kJ mol-1 Thus, ΔS° (standard entropy change ) is

Solution:

n = 4 (four electrons are involved) as

QUESTION: 11

For the reaction ,

Thus , for the reaction

Solution:

Electron involved are different, hence

Signs are taken as per required reaction

III is obtained as (I) - (II)

QUESTION: 12

Consider the following equations for a cell reaction.

A+B  C+ D, E0 = x volt, Keq = K1

2A +2B  2C+ 2D, E0 = y volt, Keq = K2

Then,

Solution:

When a chemical reaction is m ultiplied/divided EMF of the changed equation remains constant but equilibrium constant is raised to power of that change.

QUESTION: 13

Which of the following statements about the spontaneous reaction occurring in a galvanic cell is always true?

Solution:

For a spontaneous reaction,

ΔG < 0

Also,  ΔG = -nFEcell
∴    -ve = -nFEcell
∴ Ecell > 0

Also, Ecell =

if equilibrium is reached, Ecell = 0

and Q = Keq

∴

To make

Ecell >0

Ecell > Q

QUESTION: 14

For a (Ag-Zn) button cell ,the net reaction is

Zn(s) + Ag2O(s) → ZnO(s) + 2Ag(s)

ΔG0f(Ag2O) = -11.21kJmol-1,  ΔGf(ZnO) = - 318.3 kJ mol-1

Hence, E°cell of the button cell is

Solution:

ΔG0f(element, as Zn, Ag) = 0

Thus, ΔG(Button cell)

=ΔG(ZnO) - ΔGf (Ag2O)

=-318.30 - (-11.21)

=-307.09 kJ mol-1

ΔG0 = -nFE0cell        (n=2)

∴

QUESTION: 15

For the reaction ,

Thus

Solution:

for the given reaction

E0cell  = 2.73 V

n(electrons exchanged) = 12

(4 Al → Al3+ + 12e-)

F=96500 C mol-1

ΔG0 (element) = 0

∴ ΔG(reaction) = -nFE0cell

= -12 x 96500 x 2.73

=3161340 J = 3161.34 kJ

thus,

QUESTION: 16

Matching List Type

Choices for the correct combination of elements from Column I and Column II are given as options (a), (b), (c) and (d), out of which one is correct

Q.

The standard reduction potential data at 298 K is given below:

Match E° of a redox pair in Column I with the values given in Column II and select the corect answer using the codes given below:

Solution:

In all cases, we use

Equal number of electrons are involved.

QUESTION: 17

The standard potential of the following cell is 0.23 V at 288 K and 0.21 V at 308 K.

Match the parameters in Column I with their values in Column II and select the answer from the codes given below the list.

Solution:

(i) Temperature coefficient of EMF

QUESTION: 18

Comprehension Type

This section contains a passage describing theory, experiments, data, etc. Two questions related to the paragraph have been given. Each question has only one correct answer out of the given 4 options (a), (b), (c) and (d)

Passage I

The concentration of potassium ions inside a biological cell is at least twenty times higher than the outside. The resulting potential difference across the cell is important in several process such as transmission of nerve impulses and maintaining the ion balance. A simple model for such a concentration cell involving a metal M is

M(s) | M+ (aq, 0.05M), || M+(aq) 1M|M(s) | Ecell | = 77mV

Q.

For the above cell,

Solution:

QUESTION: 19

Passage I

The concentration of potassium ions inside a biological cell is at least twenty times higher than the outside. The resulting potential difference across the cell is important in several process such as transmission of nerve impulses and maintaining the ion balance. A simple model for such a concentration cell involving a metal M is

M(s) | M+ (aq, 0.05M), || M+(aq) 1M|M(s) | Ecell | = 77mV

Q.

If 0.05 M solution of M+ is replaced by 0.0025 M solution of M+,then |Ecell | would be

Solution:

QUESTION: 20

Passage II

Given,

ΔG0f(AgCl) = -109kJmol-1, ΔG0f(Cl-) = -129kJmol-1

ΔG0f(Ag+) = 77kJmol-1,

Thus E°cell of the cell reaction is

Ag+(aq) + Cl-(aq) → AgCl(s) is

Solution:

QUESTION: 21

Passage II

Given,

ΔG0f(AgCl) = -109kJmol-1, ΔG0f(Cl-) = -129kJmol-1

ΔG0f(Ag+) = 77kJmol-1,

Q.

Ksp of AgCl is thus,

Solution:

For equilibrium, AgCl(s)  Ag+  + Cl-    E0 = -0.59V

-0.59 = 0.0591logKsp

∴   log Ksp = -10

∴ Ksp = 10-10

*Answer can only contain numeric values
QUESTION: 22

One Integer Value Correct Type

This section contains 3 questions, when worked out will result in an integer value from 0 to 9 (both inclusive)

Q.

A platinum electrode is immersed in a solution containing 0.1 M Fe2+ and 0.1 M Fe3+.It iscoupled with SHE.Concentration Fe3+ of increased to 0.1 M without change in [Fe2+], then the change in EMF (in centivolt) is

Solution:

*Answer can only contain numeric values
QUESTION: 23

Using Cr2O72- aqueous, solution
E0red = 1.33V and ΔG0 = -770.07 kJmol-1

What is the valency of the ion formed after reduction?

Solution:

*Answer can only contain numeric values
QUESTION: 24

Equilibriumconstant of the cell reaction,

Cu + 2Fe3+  2Fe2+ + cu2+ is y x 1014

What is the value of y?

Solution: